Solve for $k$, $ \dfrac{k + 1}{k + 3} = -\dfrac{5}{k + 3} + \dfrac{7}{2k + 6} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $k + 3$ $k + 3$ and $2k + 6$ The common denominator is $2k + 6$ To get $2k + 6$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{k + 1}{k + 3} \times \dfrac{2}{2} = \dfrac{2k + 2}{2k + 6} $ To get $2k + 6$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{5}{k + 3} \times \dfrac{2}{2} = -\dfrac{10}{2k + 6} $ The denominator of the third term is already $2k + 6$ , so we don't need to change it. This give us: $ \dfrac{2k + 2}{2k + 6} = -\dfrac{10}{2k + 6} + \dfrac{7}{2k + 6} $ If we multiply both sides of the equation by $2k + 6$ , we get: $ 2k + 2 = -10 + 7$ $ 2k + 2 = -3$ $ 2k = -5 $ $ k = -\dfrac{5}{2}$